Skip to main content
Back to Binomial Theorem
JEE Main 2024
Binomial Theorem
Binomial Theorem
Easy

Question

The value of r for which 20 C r 20 C 0 + 20 C r-1 20 C 1 + 20 C r-2 20 C 2 + . . . . .+ 20 C 0 20 C r is maximum, is

Options

Solution

Key Concept: Vandermonde's Identity and Properties of Binomial Coefficients

This problem utilizes a specific case of Vandermonde's Identity, which relates the sum of products of binomial coefficients to a single binomial coefficient. It also requires knowledge of how to find the maximum value of a binomial coefficient.

Vandermonde's Identity states that for non-negative integers n,m,rn, m, r: k=0rnCkmCrk=n+mCr\sum_{k=0}^r {^nC_k \cdot ^mC_{r-k}} = ^{n+m}C_r This identity arises from finding the coefficient of xrx^r in the product of two binomial expansions: (1+x)n(1+x)m=(1+x)n+m(1+x)^n \cdot (1+x)^m = (1+x)^{n+m}. The coefficient of xrx^r in (1+x)n+m(1+x)^{n+m} is n+mCr^{n+m}C_r. On the other hand, by multiplying the series expansions of (1+x)n(1+x)^n and (1+x)m(1+x)^m, the coefficient of xrx^r is the sum k=0rnCkmCrk\sum_{k=0}^r {^nC_k \cdot ^mC_{r-k}}.

The maximum value of a binomial coefficient NCk^NC_k occurs when kk is the integer closest to N/2N/2.

  • If NN is even, NCk^NC_k is maximum when k=N/2k = N/2.
  • If NN is odd, NCk^NC_k is maximum when k=(N1)/2k = (N-1)/2 or k=(N+1)/2k = (N+1)/2.

Step-by-Step Solution

1. Identify the structure of the given sum. The given sum is: S=20Cr20C0+20Cr120C1+20Cr220C2++20C020CrS = ^{20}C_r \cdot ^{20}C_0 + ^{20}C_{r-1} \cdot ^{20}C_1 + ^{20}C_{r-2} \cdot ^{20}C_2 + \dots + ^{20}C_0 \cdot ^{20}C_r We can write this sum using summation notation by letting the second subscript run from k=0k=0 to k=rk=r: S=k=0r20Crk20CkS = \sum_{k=0}^r {^{20}C_{r-k} \cdot ^{20}C_k} Explanation: Each term consists of a product of two binomial coefficients. The sum of the lower indices in each product is rk+k=rr-k+k=r, which is a key characteristic for applying the identity related to the coefficient of xrx^r.

2. Relate the sum to the coefficient of a binomial expansion product. Consider the product of two binomial expansions, each for (1+x)20(1+x)^{20}: (1+x)20=20C0+20C1x+20C2x2++20Ckxk++20C20x20(1+x)^{20} = ^{20}C_0 + ^{20}C_1 x + ^{20}C_2 x^2 + \dots + ^{20}C_k x^k + \dots + ^{20}C_{20} x^{20} (1+x)20=20C0+20C1x+20C2x2++20Cjxj++20C20x20(1+x)^{20} = ^{20}C_0 + ^{20}C_1 x + ^{20}C_2 x^2 + \dots + ^{20}C_j x^j + \dots + ^{20}C_{20} x^{20} When we multiply these two expansions, we are looking for the coefficient of xrx^r in the product (1+x)20(1+x)20(1+x)^{20} \cdot (1+x)^{20}. The terms contributing to xrx^r in the product are formed by multiplying a term with xkx^k from the first expansion by a term with xjx^j from the second expansion, such that k+j=rk+j=r. This means j=rkj=r-k. The coefficient of xrx^r in the product is therefore: (20C020Cr)+(20C120Cr1)+(20C220Cr2)++(20Cr20C0)\left(^{20}C_0 \cdot ^{20}C_r\right) + \left(^{20}C_1 \cdot ^{20}C_{r-1}\right) + \left(^{20}C_2 \cdot ^{20}C_{r-2}\right) + \dots + \left(^{20}C_r \cdot ^{20}C_0\right) This can be written as k=0r20Ck20Crk\sum_{k=0}^r {^{20}C_k \cdot ^{20}C_{r-k}}. Explanation: This step links the given sum directly to the established method of finding coefficients in polynomial products. The order of multiplication of binomial coefficients within each term does not affect their product, so (20Ck20Crk)\left(^{20}C_k \cdot ^{20}C_{r-k}\right) is equivalent to (20Crk20Ck)\left(^{20}C_{r-k} \cdot ^{20}C_k\right), confirming the equivalence to the given sum.

3. Simplify the expression using exponent rules. Since (1+x)20(1+x)20=(1+x)20+20=(1+x)40(1+x)^{20} \cdot (1+x)^{20} = (1+x)^{20+20} = (1+x)^{40}, the given sum is simply the coefficient of xrx^r in the expansion of (1+x)40(1+x)^{40}. Explanation: This simplification leverages the fundamental property of exponents aman=am+na^m \cdot a^n = a^{m+n} and the definition of a binomial coefficient from the binomial theorem, (1+x)N=k=0NNCkxk(1+x)^N = \sum_{k=0}^N {^N C_k x^k}.

4. Determine the value of rr for which the coefficient is maximum. From the binomial theorem, the coefficient of xrx^r in (1+x)40(1+x)^{40} is 40Cr^{40}C_r. For a binomial coefficient NCk^NC_k, its maximum value occurs when kk is equal to N/2N/2 (if NN is even) or the two integers closest to N/2N/2 (if NN is odd). In this case, N=40N=40, which is an even number. Therefore, 40Cr^{40}C_r is maximum when r=402=20r = \frac{40}{2} = 20. Explanation: The binomial coefficients exhibit symmetry, peaking at the middle term(s) of the expansion. For an even power NN, there is a unique middle term, and thus a unique maximum coefficient.


Tips and Common Mistakes

  • Recognize the pattern: The sum nCkmCrk\sum {^nC_k \cdot ^mC_{r-k}} is a common pattern in combinatorics that almost always points to Vandermonde's Identity or the coefficient of a product of binomial expansions.
  • Don't get confused by indices: Ensure that the sum of the lower indices in each product equals the target power rr.
  • Property of NCk^NC_k maximum: Remember that NCk^NC_k is maximum at k=N/2k = N/2 for even NN, and at k=(N±1)/2k = (N \pm 1)/2 for odd NN. This is a frequently tested concept.
  • Latex Formatting: Using proper LaTeX for mathematical expressions (like nCk^nC_k for combinations and ...... for display equations) is crucial for clarity in mathematics.

Summary and Key Takeaway

The problem demonstrates how a seemingly complex sum of products of binomial coefficients can be elegantly simplified by recognizing its connection to the coefficient of a term in the expansion of a single binomial power. By identifying the sum as 40Cr^{40}C_r, the problem reduces to finding the maximum value of a standard binomial coefficient, which occurs at the central term. This highlights the power of combinatorial identities and the properties of binomial coefficients.

The final answer is 20\boxed{\text{20}}.

Practice More Binomial Theorem Questions

View All Questions